Question

Find the general solution of the given differential equation. Give the largest interval / over which the general solution is defined. Determine whether there are any transient terms in the general solution.

$\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{xy}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{x}$

Short answer

So, the general solution of the given differential equation is $y=-\frac{\left(x^2+3(x+1)\right)}{1+x}+\left(\frac{e^x}{1+x}\right)$.

Step-by-step solution

Definition of transient term

A transient term means that you (or someone or something) will be moving on from where you are now.

Evaluation

Consider the differential equation $(1+x)\frac{dy}{dx}-xy=x+x^2$.

Rewrite the differential equation, divide both sides (1+x) then,

role="math" $\begin{array}{rcl}(1+x)\frac{dy}{dx}-xy& =& x+x^2\\ \frac{dy}{dx}-\left(\frac{x}{1+x}\right)y& =& \frac{x(1+x)}{1+x}\\ \frac{dy}{dx}+\left(-\frac{x}{1+x}\right)y& =& x\end{array}$

This implies,

$\frac{dy}{dx}+\left(-\frac{x}{1+x}\right)y=x\dots \dots .\left(1\right)$

Compare the differential equation (1) with the standard form of the first-order linear differential $\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)$.

Here $P\left(x\right)=-\left(\frac{x}{1+x}\right)andf\left(x\right)=x$

Seek a solution of equation (1) on an interval l for which both coefficient functions l and f are continuous.

As $P\left(x\right)=-\left(\frac{x}{1+x}\right)$and $f\left(x\right)=x$, observe that the functions and are continuous on $(-\infty ,\infty )$expect X = -1.

Use Substitution method to integrate

Now, calculate the integrating factor $e^{\int P\left(x\right)dx}$.

Substitute $P\left(x\right)=-\left(\frac{x}{1+x}\right)$and evaluate it,

$$\begin{gathered} e^{\int -\left(\frac{x}{1+x}\right)dx}=e^{-\int \frac{1+x-1}{1+x}dx} \\ =e^{\ln (1+x)-x} \end{gathered}$$

And

$$\begin{gathered} e^{\int -\left(\frac{x}{1+x}\right)dx}=e^{\ln (1+x)}{e}^{-x} \\ =(1+x)e^{-x} \end{gathered}$$

Multiply equation (1) by integrating factor provides us:

$(1+x)e^{-x}\frac{dy}{dx}-x{e}^{-x}y=x{e}^{-x}+x^2{e}^{-x}$

The $-x{e}^{-x}$term is written as $\frac{d}{dx}\left((1+x)e^{-x}\right)$.

The L.H.S term is the expend form of the differential of the two functions as:

$\frac{d}{dx}\left(f\right(x\left)g\right(x\left)\right)=f\left(x\right)\frac{d}{dx}g\left(x\right)+g\left(x\right)\frac{d}{dx}f\left(x\right)$

So, the above L.H.S expression is written as:

$\frac{d}{dx}\left[y(1+x)e^{-x}\right]=x{e}^{-x}+x^2{e}^{-x}$

Integrate on both sides of the equation with respect to X

$$\begin{gathered} \int \frac{d}{dx}\left[y(1+x)e^{-x}\right]dx= \\ \int \left(x{e}^{-x}+x^2{e}^{-x}\right)dx \\ \left[x\int e^{-x}dx-\int \left[\frac{d}{dx}\left(x\right)\int e^{-x}dx\right]\right] \\ dx+x^2\int e^{-x}dx-\int \left[\frac{d}{dx}\left(x^2\right)\int e^{-x}dx\right]dx \\ =\left[x\left(-e^{-x}\right)-\int \left[1·\left(-e^{-x}\right)\right]dx+x^2\left(-e^{-x}\right)\right] \\ -\int \left[2x·\left(-e^{-x}\right)\right]dx \end{gathered}$$

This implies,

$$\begin{gathered} =-x{e}^{-x}-e^{-x}-x^2{e}^{-x}+2\left(-x{e}^{-x}-e^{-x}\right)+c \\ =-e^{-x}\left(x^2+3(x+1)\right)+c \end{gathered}$$

Solving for

$$\begin{gathered} y(1+x)e^{-x}=-e^{-x}\left(x^2+3(x+1)\right)+c \\ y(1+x)=-\frac{e^{-x}\left(x^2+3(x+1)\right)}{e^{-x}}+\frac{c}{e^{-x}} \\ y=-\left(x^2+3(x+1)\right)+c{e}^x \\ -\frac{\left(x^2+3(x+1)\right)}{1+x}+\left(\frac{e^x}{1+x}\right) \end{gathered}$$

Hence, the required solution is $y=-\frac{\left(x^2+3(x+1)\right)}{1+x}+\left(\frac{e^x}{1+x}\right)$for $-\infty <x<\infty$except x = -1.

The term is said to be a transient if its value tends to zero as the independent variable tends to infinity. As $x\to \infty$the term $e^x\to \infty and{e}^{-x}\to 0$.

Thus, there is no transient term.